F(x)=8th root of (8-x^2)sec(4x)... cant figure out the derivitive of this.. any help?

Question

anonymous · Answer

try splitting it up into two parts: [(8-x^2)^(1/8)][(sec(4x))^(1/8)] Sorry if all those parens get confusing, but basically its the two seperate parts being raised to the 1/8 power and are multiplied.

anonymous · Answer

sorry the sec(4x) isnt in the root. its being mulitplied by the 8th root of (8-x^2)

anonymous · Answer

Well either way, its still just f(x) = ab, so $$\prime$$(x) = (a)(db) + (b)(da)

anonymous · Answer

$$f \prime(x)*$$

anonymous · Answer

what would the der. of the 8th root of (8-x^2) be? i got how the formulas to use are, but i dont know what im doing wrong here...

anonymous · Answer

Thats the chain rule. So it would be (1/8)*[(8-x^2)^(-7/8)]*(-2x)

anonymous · Answer

ok i see where i went wrong.. thanks a lot